Question 3: Sequences from the Alphabet.
a) What are the next three terms for the following sequences?
i) 8, 5, 4, 9, 1, 7, 6, . . .
ii) o, t, t, f, f, s, s, . . .
iii) 4, 8, 12, 2, 1, 7, 6, . . .
b) What is the next number in this pattern?
2, 4, 6, 30, 32, 34, 36, 40, 42, 44, 46, 50, 52, 54, 56, 60, 62, 64, 66, ...
Answer
a)
i) 3, 2, 0 (the first ten numbers (from 0 to 9) in alphabetical order: eight, five, four, nine, etc.)
ii) e, n, t (the initials of the first ten counting numbers: one, two, three, etc.)
iii) 3, 5, 11 (the months of the year in alphabetical order)
b) 2000, as all of these numbers do not contain the letter E, when being spelt (in English).
Question 4: More simple problems:
a) How many different 4digit numbers can be made from the digits {1, 2, 3} if each digit must appear at least once?
b) A painted rectangular prism with sides 4 cm, 5 cm, and 6 cm is cut up into 1 cm cubes. How many of these cubes are not painted at all ?
c) In a room of 100 people if everyone has to shake hands with everyone in the room once, how many handshakes are made altogether?
Answer
a) 36: Consider a 3digit number formed from {1, 2, 3}, there are 3×2×1 = 6 ways, the fourth digit can be any of the 3 numbers and it can be at any of the 4 places. Total = 6×3×4 = 72 ways. But since there are two same digits in each number, dividing 72 by 2 gives 36 numbers.
b) 24: Assume there is a layer of 1 cm around the cube. Take the layer off, the new cube has dimensions 2×3×4, i.e. 24 cubes.
c) Two people have 1 handshake, 3 people have 3 handshakes, 4 people have 6 handshakes, . . ., the formula is n(n – 1)/2.
For 100 people, there are 100 × 99 ÷ 2 = 4950 handshakes.
Question 5: Mysterious ages
Two women A and B are talking.
A: I have three daughters whose ages are found by the following clues. Stop me as soon as you know their ages. Clue 1: The sum of their ages is 13.
B: I need more clues.
A: Clue 2: The product of their age is the same as your age.
B: I need more clues.
C: Clue 3: My oldest daughter is sleeping upstairs.
B: Stop. I know their ages.
Do YOU know her three daughters' ages?
Answer
From the first clue, their ages can be one of the following sets (1, 1, 11), (1, 2, 10), (1, 3, 9), (1, 4, 8), (1, 5, 7), (1, 6, 6), (2, 2, 9), (2, 3, 8), (2, 4, 7), (2, 5, 6), (3, 3, 7), (3, 4, 6), (3, 5, 5) and (4, 4, 5). Their respective products are 11, 20, 27, 32, 35, 36, 36, 48, 56, 60, 63, 72, 75 and 80, of which one is B's age. Of course you do not know B's age but she must know her own age. The only reason that she needs one more clue is there are two sets of ages that give the same product (= her age). Here, both (1, 6, 6) and (2, 2, 9) give the product of 36. The third clue indicates that A has an oldest daughter. So, their ages are 2, 2, and 9.
Question 6: The Missing Dollar
Three men shared a room in a hotel and were charged $30 for the night. They all agreed to share the cost, so each paid $10. The clerk later found that he had overcharged them by $5 so he asked a waiter to return them $5. The waiter reasoned that it was not easy to divide the $5 by three, so he pocketed $2 and returned only $3.
Now, as each man paid $9 apiece, or a total of $27 for the room. Add that to the $2 the waiter kept gives the total of $29. Where did the missing dollar go?
Answer
The three men paid $30 originally, they were refunded $3 so they paid $27 for the room. The waiter kept $2 from this $27, so correctly we must not add $2 with $27, rather it should be subtracted: $27 less $2 gives $25 (The men paid $27, the manager kept $25, the waiter pocketed $2, everything now is correct!).
Question 7: A piece of cake!
Can you cut a cake into 14 pieces with four straight cuts?
Answer
If you draw your own diagram as you read, it would be easier to understand this: 2 cuts give 4 slices. Make the third cut through the previous two cuts then you have 3 more pieces: Total 7 pieces by far. Make the fourth cut through the other three cuts then you have 4 more pieces: Total 11 pieces at most.
However, if you think of the cake as a cylinder, then make the fourth cut parallel with the top surface of the cake: You double the number of slices you have had after three cuts. Total = 7 × 2 = 14 slices!
Question 8: How many squares in a chessboard?
How many squares are there in the 8 squares by 8 squares chessboard?
Answer
64 1x1 squares, 49 2x2 squares, 36 3x3 squares, 25 4x4 squares, 16 5x5 squares, 9 6x6 squares, 4 7x7 squares and 1 8x8 square. Total = 204 squares.
[Harder question: How many rectangles are there in the 8 squares by 8 squares chessboard?
Answer: A rectangle is formed by two vertical grid lines and two horizontal grid lines. Since a chessboard has 9 vertical lines and 9 horizontal lines, we can find the number of rectangles by choosing 2 of the 9 lines, both vertically and horizontally. Therefore, the answer is 9C2 X 9C2 = 1296.]
Question 9: How many triangles?
With 6 match sticks how many equilateral triangles of side equal to the length of a match stick can you make at most?
Answer
4, if you build the frame of a tetrahedron (a triangular pyramid)
Question 10: Locker problem
In a hall there were 50 lockers that belonged to 50 year twelve students. On the year twelve muckup day, the 50 students decided to go through the hall one after another and played this trick:
The first student shut all the lockers. The second student opened every second locker. The third student changed every third locker (i.e. if it was opened, he/she closed it; and if it was closed he/she opened it). The fourth student changed every fourth locker, and so on. How many lockers remained closed after all 50 students had a go?
Answer
Consider the 12th locker, for example, it was shut by the 1st student, opened by the 2nd student, shut by the 3rd student, opened by the 4th student, shut by the 6th student, opened by the 12th student. These numbers {1, 2, 3, 4, 6, 12} are the factors of 12. So, the principle is: if a locker has an even number of factors it would be opened at the end, and of course if a locker has an odd number of factors it would be closed at the end. Only {1, 4, 9, 16, 25, 36, 49} have an odd number of factors. Therefore, at the end 7 lockers were shut.
Question 11: Successive savings
An homeowner makes three successive designs that save, in turn, 20%, 55%, and 25% on the heating costs of the house. Find the overall percentage saved.
Answer
73%: Since 80%×45%×75% gives 27%; a saving of 73%.
Question 12: A clock problem
How many times, not in 24 hour format, can digital clock displays in half a day are perfect square ? (A display such as 12:22 reads as 1222)
Answer
16: The minimum display is 100 (i.e. 1:00) and the maximum is 1259 (i.e. 12:59). The least perfect square is 10² = 100 and the greatest is 35² = 1225, so there are 35 – 10 + 1 = 26 perfect square numbers.
However, as the last two digits cannot be greater than 60, these 10 numbers must be excluded (13, 14, 17, 19, 22, 24, 26, 28, 31, 33 because 13² = 169, 14² = 196, 17² = 289, 19² = 361, 22² = 484, 24² = 576, 26² = 676, 28² = 784, 31² = 961, 33² = 1089).
(Thanks to Edwin Cooper for his solution)
Question 13: Double vision
A water lily doubles itself in area each day. After 12 days the whole pool is completely covered by the water lily. When was the pool half covered?
Answer
11 days.
Question 14: Numbers matter
The value of a threedigit number is twelve times the sum of its digits. What is the number?
Answer
Let abc be the number, then 100a + 10b + c = 12(a + b + c), or by simplifying, 88a = 2b + 11c.
Only one equation but three unknowns! Impossible? Wait, as a, b, c have to be single digits, we can solve it by trial and errors.
For a = 1, LHS = 88. The only solution is when b = 0, 88 = 11c, so a = 1, c = 8. The number is 108.
For a = 2 or higher, LHS = 176 or higher, while the maximum value for the RHS is 2×9 + 11×9 = 117: There are no other values of a, b, c satisfying the given condition. Thus, the only solution is 108.
Question 15: Crossing the river
Three couples come to a river. There they find a small boat that can carry only two people at a time. How can all six people cross the river if no wife is allowed to be left with any man at any stage without her husband's presence. Assume that anyone can row.
[If you find this one easy, try the next one]
Answer
Let the men be A, B, C and their wives be a, b, c respectively.
This side of river

Forward trip

Return trip

The other side

A, a, B, b, C, c

A, a

A

a

A, B, b, C, c

b, c

a

b, c

A, a, B, C

B, C

B, b

C, c

A, a, B, b

A, B

c

A, B, C

a, b, c

a, b

C

A, a, B, b

C, c

C, c

NIL

A, a, B, b, C, c

Question 16: Crossing the river again
Three couples come to a river. There they find a small boat that can carry only two people at a time. How can all six people cross the river if:
Rule 1: The number of women must be equal or more than the number of men at any stage;
Rule 2: Of the six people, only one man and his wife can row the boat.
Answer
Let the men be A, B, C and their wives be a, b, c respectively. Assume A and a can row.
This side of river

Forward trip

Return trip

The other side

A, a, B, b, C, c

A, B

A

B

A, a, b, C, c

A, C

A

B, C

A, a, b, c

a, b

a, C

B, b

A, a, C, c

A, a

a, B

A, b

a, B, C, c

a, c

A

a, b, c

A, B, C

A, B

A

a, B, b, c

A, C

A, C

NIL

A, a, B, b, C, c

Question 17: Time traveller
What is the greatest number of Mondays which can occur in a 45day period?
Answer
45 ÷ 7 = 6, remainder = 3. Thus, if Monday is the 1st, 2nd or 3rd day of the 45day period then there will be 7 Mondays within the period.
Question 18: Jack, Jill and a jug
Jack and Jill each have a 1 litre jug of water. The first day, Jack pours 1 mL of water from his jug into Jill's. The second day Jill pours 3 mL of water from her jug into Jack's. The third day Jack pours 5 mL of water from his jug into Jill's, and so on. What is the volume of water in Jack's jug at the end of the 101st day?
Answer
The volumes of water in Jack's jug are:
Day 1: 999 mL, day 3: 997 mL, day 5: 995 mL, and so on. This is an AP, with a = 999, d = – 2 so on day 101 (n = 51) the amount of water = a + (n – 1)d = 999 + 50 × –2 = 899 mL.
Question 19: A 2KY problem
How many years in the 21st century will have the property that, dividing their year number by each of 2, 3, 5 and 7 always leaves a remainder of 1?
Answer
To divide by each of 2, 3, 5 and 7 with a remainder of 1 the number less 1 must be a multiple of 2 × 3 × 5 × 7, i.e. in the form 210k + 1.
Put k = 9 gives 1891, put k = 10 gives 2101. Thus, for the years from 2000 to 2100 there is none that satisfies the given condition.
Question 20: A triangle problem
A triangle ABC is to be formed having all the sides of integer length. If AB = 37 and AC = m, where m is a fixed integer less than 37, how many different lengths are possible for BC?
Answer
This problem can be viewed geometrically: Draw a line AB of length 37 units. At A draw a circle of radius m units, m < 37, and ask yourself how many circles of integer radius can be drawn at B to intersect the first circle. Obviously the smallest radius = 38 – m, and the largest radius = 36 + m. Thus, the number of circles = (36 + m) – (38 – m) + 1 = 2m – 1.
Question 21: Another triangle problem
If we are given a square we can easily construct a triangle enclosing it as shown (i.e. one side of the square lies on a side of the triangle, the other vertices of the square lie on the other sides of the triangle). Obviously there are many such triangles. However, if we are given a triangle how can you construct such a square?
Answer
Let the given triangle's height and base be h units and a units respectively. Assume such a square is drawn inside the triangle, let the side of the square be x units. Also, let its corner N divide the side AC in the ratio p : q
ABC is similar to AMN, so a/x = (p + q)/p = 1 + q/p. Thus, q/p = a/x – 1.
CAH is similar to CNP, so h/x = (p + q)/q = 1 + p/q. Thus, p/q = h/x – 1.
Multiplying these two equations gives 1 = (a/x – 1)(h/x – 1).
Thus, ah/x² – (a + h)/x = 0, on expanding then simplifying
Thus, x = ah/(a + h).
That is, given a triangle of base a units, height h units the square can be drawn with side MN = NP = ah/(a + h).
The following riddles were given to the 40 students attending the EndOfYear Cruise (2009).
1
(i) I am the beginning of the end, and the end of the time. I am essential to the creation of the universe, and I surround everyone. What am I?
(ii) What is in seconds, but not in hours; in months but not in years, and in centuries but not in decades?
2
On a boat of 40 people, if everyone has to shake hands with everyone in the room once, how many handshakes are made altogether?
3
Solve this equation: x to the power of x to the power of x to the power of x, indefinitely = 2.
4
When I taught at Newington College, in my first day of a year 10 Math class I noticed two almost identical looking boys sitting in the front row.
To break the ice, I asked: "Are you twins?".
Immediately they responded in unison: "No".
The class burst out laughing. But later I found out that they have the same parents and were born on the same day. How could it be?
5
Two students arrived late for their examination. They entered the classroom and solemnly told their teacher they had been detained due to a flat tyre. The sympathetic teacher told them that she was willing to let them make it up. So she sent them to two different rooms where they had to answer just one question. The question was ‘Which tyre?’
What is the probability that their answers match each other, assuming they lied about the flat tyre and they were unprepared for this question?
6
In a hall there were 50 lockers that belonged to 50 year twelve students. On the year twelve muckup day, the 50 students decided to go through the hall one after another and played this trick: The first student shut all the lockers. The second student opened every second locker. The third student changed every third locker (i.e. if it was opened, he/she closed it; and if it was closed he/she opened it). The fourth student changed every fourth locker, and so on. How many lockers remained closed after all 50 students had a go?
7
Recite pi correct to 10 decimal places.
8
One hundred people lined up to board an airplane, but the first passenger lost his boarding pass (which shows his seat number), therefore he just took a random seat. What is the probability that the last passenger to board finds his seat occupied by the first passenger?
9
(i) Differentiate (ax+b)/(cx+d)
(ii) Integrate 2x/sqrt(1 + x^2)
10
I bought a poorly designed analog clock that has identical minute and hour hands. How many times in a day can the time be told exactly using only the clock?
11
When all the counting numbers from 1 to 100 inclusive are written down, how many digits are used and which digit appears the most and also which digit appears the least?
12
(i) Which English word that when the 1st, 3rd, and 5th letters are removed is still pronounced by the same way?
(ii) Divide 30 by half and add 10.
Answer:
1) (i) letter E, (ii) letter N.
2) 39 X 20 = 780, as 1 student has no handshake, 2 students have 1 handshake, 3 students have 3 handshakes, etc. The pattern is (n  1)(n)/2 for n students.
3) sqrt 2. Because the powers of x are also "x to the power of x to the power of x to the power of x, indefinitely", which is 2 (infinity less 1 is also infinity), so basically you are required to solve x^2 = 2.
4) Triplet. The third one is a girl, so she does not go to a boys school.
5) In theory the answer is 4C1(1/4 X 1/4) = 1/4. However, people do not choose the answer randomly, they reason before they do. According to Ask Marilyn, (www.marilynvossavant.com), 58% responded that they would choose the front right tyre in her survey.
6) 7. This question is the same as the locker problem (question 10) above.
7) 3.1415926535 (Can I find a trick recalling pi easily? Count the words).
8) 0.5, as either the seat is taken or not taken by the first passenger.
9) (i) (ad  bc)/(cx + d)^2, (ii) 2 sqrt (1 + x^2) + C.
10) 22, as the time can be told exactly only when the two hands coincide. The minute hand rotates 24 times in a day, so if the hour hand stops, they meet each other 24 times. But the hour hand rotates twice in a day (by moving forward), therefore, they meet each other 22 times.
11) 192 digits are used, 1 is the most, 0 is the least, as there are 9 single digit numbers, 2 X 9 X 10 = 180 double digit numbers, and 3 X 1 triple digit numbers. Total = 192, of which 2, 3, ..., 9 are used 20 times, 1 is used 21 times, and 0 is used 11 times.
12) (i) EMPTY, as once the letters E, P, Y are removed, you are left with MT, (ii) 70, as 30/(0.5) + 10 = 60 + 10 = 70.
